Induced map in k-theory by an involution

Question

Let $T$ be a ring with involution $s:T\rightarrow T$. And let $$h:T\otimes T^\text{op} \rightarrow T\otimes T^\text{op}$$ be the ring automorphism given by $h(a\otimes b)=s(b)\otimes s(a)$.

suppose that $$ K_{0}(h): K_{0}(T\otimes T^\text{op})\rightarrow K_{0}(T\otimes T^\text{op}) $$ is the identity map.

I was wondering if the induced homorphism $$ K_{\ast}(h): K_{\ast}(T\otimes T^\text{op})\rightarrow K_{\ast}(T\otimes T^\text{op}) $$ in K-theory, is the identity map?

Notice that "op" is used for the opposite multiplication.

Answer 1

I don't think so: Let us take $T=\mathbb{Z}[x^{\pm 1}]$, and let us take $s$ the identity. Then $T\otimes T^{op}= \mathbb{Z}[x^{\pm 1},y^{\pm 1}]$ has $K_0 = \mathbb{Z}$ detected by rank (by Grothendieck-Serre), so $K_0(h)$ is the identity. But I think in $K_1$ the elements given by the units $x$ and $y$ are different, and are interchanged by $K_1(h)$.